In an $LCR$ circuit,the inductance is changed from $L$ to $9 L$. For the same resonant frequency,the capacitance should be changed from $C$ to:

For an $L-C-R$ $AC$ circuit,the resonance frequency is $5000 \ Hz$ and the frequencies at half-power points are $4950 \ Hz$ and $5050 \ Hz$. What will be the $Q$-factor?

The reading of the ammeter in the following circuit is (in $A$)

$A$ parallel combination of a pure inductor and a capacitor is connected across a source of alternating e.m.f. '$e$'. The currents flowing through the inductor and the capacitor are $i_{L}$ and $i_{C}$ respectively. In this parallel resonant circuit,the condition for the currents $i$,$i_{L}$,and $i_{C}$ is ($i =$ net r.m.s. current in the circuit).

An $LCR$ circuit is at resonance for a capacitor $C$,inductance $L$,and resistance $R$. If the value of the resistance is halved while keeping all other parameters the same,the current amplitude at resonance will be:

In a series resonant $R-L-C$ circuit, the voltage across $R$ is $100 \, V$ and the value of $R = 1000 \, \Omega$. The capacitance of the capacitor is $2 \times 10^{-6} \, F$; the angular frequency of the $AC$ source is $200 \, rad \, s^{-1}$. Then the potential difference across the inductance coil is: (in $V$)